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Behavior of Gases
![](3D"slide0001_image004.gif")
![](3D"master04_background.jpg")
•Describe the properties of gas particles.
•Explain how the kinetic energy of gas particles re=
lates
to Kelvin
temperature.
•Explain how the amount of gas and the volume of th=
e container affect gas pressure.
<=
/div>
• Infer the effect of temperature changes on the pre=
ssure exerted by a contained gas.
• Distinguish between ideal and real gases.
• State Avogadro’s hypothesis, Dalton’s =
Law,
and Graham’s Law.=
3;
<=
span
style=3D'font-size:69%'>•Calculate the amount of gas at any specified condi=
tions
of pressure, volume=
and
temperature.
<=
span
style=3D'font-size:69%'>•Calculate moles, masses, and volumes of gases at
STP.
•Calculate poartial pressures and rates of
effusiojn.
TEKS:
2A, 2B, 2C, 2D, 2E, 3B, 3D, 3E, 4B, 4C, 7A, 7B, 11C
Kinet=
ic
Theory
<=
span
style=3D'font-size:86%'>–Gas – composed of particles (molecules or
atoms)
<=
span
style=3D'font-size:86%'>–Particles move in constant random motion
<=
span
style=3D'font-size:86%'>–All collision are perfectly elastic.
<=
span
style=3D'font-size:75%'>•Gas particles are so small in the relationship to =
the distances between them that their individual volum=
es can be considered insignificant
=
<=
span
style=3D'font-size:75%'>•Gas particles move rapidly in random motion.
<=
/span>
Boyle=
’s
Law
•States that for a given mass of gas at a constant =
temperature, the volume of the gas varies inversel=
y with pressure.
P1<=
/span>
x V1<=
/span>
=3D P2<=
/span>
x V2&=
#13;
•#10
The pressure on 2.50 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will the new volume if the temperature remains constant?
•#10
The pressure on 2.50 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will the new volume if the temperature remains constant?=
3;
=
;
P1 x V1
=3D P2<=
/span>
x V2&=
#13;
=
;
P2
=
;
V2 =3D
2.50 L x 105 k Pa =3D
6.48 L
=
;
40.5 k Pa
![](3D"slide0061_image005.gif")
![](3D"slide0061_image006.gif")
•#11 =
A gas
with a volume of 4.00 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L=
. What is the pressure in the container if the temperature remains constant?
P1 x V1
=3D P2<=
/span>
x V2&=
#13;
•
P2 =3D
P1 x V1
=
;
V2
•
P2 =3D 205 k Pa x 4.00 L =3D
68.3 k Pa
=
; &n=
bsp;
12.0 L
•
![](3D"slide0062_image007.gif")
![](3D"slide0062_image008.gif")
Charl=
es’
Law
•States that the volume of a fixed mass of gas is <=
/span> directly proportional to its Kelvin temperature if=
the pressure is kept constant. Temperature must be measured in Kelvin. K =3D
&or=
dm;C +
273
V1<=
/span> =
;
=3D =
V2&=
#13;
T1<=
/span> =
;
T2
![](3D"slide0040_image009.gif")
•#12 =
If a
sample of gas occupies 6.80 L at 325 ºC, =
what will be its volume at 25 ºC if the
pressure does not c=
hange?
<=
span
style=3D'font-size:75%'>•#12 If a sample of gas occupies 6.80 =
L at
325 º=
C, what
will be its volume at 25 ºC if the pressure does not change?
=
; &n=
bsp;
V1 =3D V<=
span
style=3D'font-size:50%;position:relative;top:.37em;mso-text-raise:-25%'>2&=
#13;
• T1<=
/span> =
;
T2
•
• =
;
325 °C + 273 =3D
598 K
• 25 =
°=
C + =
273 =3D
298 K
•
• V<=
span
style=3D'font-size:50%;position:relative;top:.37em;mso-text-raise:-25%'>2<=
/span> =3D=
V1 x T2
• =
;
T1
•
• V<=
span
style=3D'font-size:50%;position:relative;top:.37em;mso-text-raise:-25%'>2<=
/span> =3D=
6.80 L x
298 K =3D 3.39 L
• =
; &n=
bsp;
598 K
•
•
![](3D"slide0063_image010.gif")
![](3D"slide0063_image011.gif")
![](3D"slide0063_image012.gif")
•#13 =
Exactly
5.00 L of air at -50.0 ºC is warmed to <=
/span>100.0 ºC. What is the new volume if the pre=
ssure remains constant?
•#13 =
Exactly
5.00 L of air at -50.0 ºC is warmed to <=
/span>100.0 ºC. What is the new volume if the pre=
ssure remains constant?
=
; &n=
bsp;
V1 =3D
V2
=
; &n=
bsp;
T1 T<=
span
style=3D'font-size:50%;position:relative;top:.37em;mso-text-raise:-25%'>2&=
#13;
V<=
span
style=3D'font-size:50%;position:relative;top:.37em;mso-text-raise:-25%'>2<=
/span> =3D=
V1 x T2
=
;
T1
V<=
span
style=3D'font-size:50%;position:relative;top:.37em;mso-text-raise:-25%'>2<=
/span> =3D=
5.00 L x
373 K =3D 8.36 L
=
; &n=
bsp;
223 K
![](3D"slide0065_image013.gif")
![](3D"slide0065_image014.gif")
![](3D"slide0065_image015.gif")
Gay
Lussac’s Law
•States that the pressure of a gas is directly proportional to the Kelvin temperature if the volu=
me is kept constant.&nb=
sp;
Temperature must be measured in Kelvin =
;
K =3D ºC + 273
• P1<=
/span>
=3D P2&=
#13;
• T1<=
/span> =
;
T2
![](3D"slide0043_image016.gif")
•#14 =
A gas
has a pressure of 6.58 kPa at 539 K.
What will be the
pressure at 211 K if the volume does not
change?
•#14 =
A gas
has a pressure of 6.58 kPa at 539 K.
What will be the
pressure at 211 K if the volume does not
change?
=
; &n=
bsp;
P1 =3D
P2
=
; &n=
bsp;
T1 T<=
span
style=3D'font-size:50%;position:relative;top:.37em;mso-text-raise:-25%'>2&=
#13;
P1 =3D
P2 x
T1
=
;
T2
P1 =3D
6.58 k Pa x 211 K =3D 2.58 k Pa
=
; &n=
bsp;
539 K Pa
![](3D"slide0066_image017.gif")
![](3D"slide0066_image018.gif")
![](3D"slide0066_image019.gif")
•#15 =
The
pressure in an automobile tire is 198 kPa at 27 ºC. At the end of a trip on a hot sun=
ny
day, the pressure has ris=
en to
225 kPa. What is the =
temperature of the air in the tire? (Assume that the volume has not changed.)
•#15 =
The
pressure in an automobile tire is 198 kPa at 27 ºC. At the end of a trip on a hot sun=
ny
day, the pressure has ris=
en to
225 kPa. What is the =
temperature of the air in the tire? (Assume that the volume has not changed.)
=
; &n=
bsp;
P1 =3D P2
=
; &n=
bsp;
T1 T2&=
#13;
=
;
T2 =3D P2 x T=
1&=
#13;
=
;
P1
=
;
T2 =3D
225 k Pa x 300 K =3D 340.9 =3D
340 K
=
; &n=
bsp;
198 k Pa
![](3D"slide0067_image020.gif")
![](3D"slide0067_image021.gif")
![](3D"slide0067_image022.gif")
Combi=
ned
Gas Law
• P1 x V1 =
;
=3D P2<=
/span>
x V2&=
#13;
• T1<=
/span> =
;
T2
![](3D"slide0045_image023.gif")
•#16 =
A gas at
155 kPa and 25 ºC occupies a container
with an initial volume of 1.00 L.
By changing the vol=
ume,
the pressure of the gas increase
to 605 kPa as the temperature is raised to 125 ºC. What is the new volume?
=
; &n=
bsp;
•#16 =
A gas at
155 kPa and 25 ºC occupies a container
with an initial volume of 1.00 L.
By changing the vol=
ume,
the pressure of the gas increase
to 605 kPa as the temperature is raised to 125 ºC. What is the new volume?
=
; &n=
bsp;
P1 x V1 =
;
=3D P2<=
/span>
x V2&=
#13;
=
;
T1 =
;
T2
V2 =3D
P1 x
V1 x
T2
=
;
T1 x
P2
V2 =3D
155 k Pa x 1.0 L x 398
K =3D 0.34 L
=
;
298 K x 605 k Pa
![](3D"slide0046_image024.gif")
![](3D"slide0046_image025.gif")
![](3D"slide0046_image026.gif")
Ideal=
Gas
Law
• P x V =3D n x R x T
• R Values :
0.0821 L x atm
• K x mol
•
62.4 L x mmHg
K x mol
8.31 L x kPa
K x mol
![](3D"slide0047_image027.gif")
![](3D"slide0047_image028.gif")
![](3D"slide0047_image029.gif")
•#24 =
A child
has a lung capacity of 2.20 L. How
many grams of air do her lungs hold at a pre=
ssure
of 102 kPa and a no=
rmal
body temperature of 37ºC? Air is a mixture, but you may assume an average molar mass of 29 g/mol for air because air is abou=
t 20% O2 , (m=
olar
mass 32) and 80% N2 (molar mass 28).
•#24 =
A child
has a lung capacity of 2.20 L. How
many grams of air do her lungs hold at a pre=
ssure
of 102 kPa and a normal body
temperature of 37ºC? Air is a mixture, <=
span
style=3D'position:absolute;top:50.39%;left:3.52%;width:95.06%;height:16.53=
%'>but you may assume an average molar mass of 29 g/m=
ol for air because air is about 20% O2 =
, (molar mass 32) and 80% N2 (molar mass 28).
P x V =3D n
x R x
T
n =3D P
x V
=
;
R x T
=
;
n =3D 1 atm x
2.20 L
=
;
0.0821 L atm
x 310 K
=
; &n=
bsp;
K x mol
=
;
n =3D 0.086 mol of air
=
;
0.086 mol air x 29.0 g/mol &n=
bsp;
=3D 2.5 g of air
![](3D"slide0069_image030.gif")
![](3D"slide0069_image031.gif")
![](3D"slide0069_image032.gif")
Ideal=
Gas Law
Avoga=
dro’s
Hypothesis
•Equal volumes of gas at same temperature and same =
pressure contain equal number of particles.
• 1 mol =3D 6.02 x 1023=
particles
=
•#31 =
What is
the volume occupied by 0.250 mol of a gas at STP?
•#33 =
How many
nitrogen molecules are in 5.12 L of the
gas at STP?
=
• 0.250 mol gas x 22.4 L gas =3D
5.60 L gas
=
=
; &n=
bsp; =
;
1 mol gas
•
•
•
•
=
=
;
5.12 L gas x 1 mol gas x 6.02 x 1023 molec
gas
=
=
; &n=
bsp;
22.4 L gas 1 m=
ol
gas
=
=3D 1.78 x 1023=
mole=
cules
nitrogen
![](3D"slide0054_image033.gif")
![](3D"slide0054_image034.gif")
•#35 =
What is
the volume of a container that hold 8.80g
of CO2 at STP ?
•
8.80 g CO2 x
1 mol CO2 x
22 .4 L =3D=
4.48 L
=
; &n=
bsp;
44.0 g CO2 1 mol CO2&=
#13;
![](3D"slide0071_image035.gif")
Dalto=
n’s
Law
•At a constant volume and temperature, the total pressure exerted by a mixture of gases is equal to=
the sum of the partial pressures of the component =
gas.
PT<=
/span>
=3D P1<=
/span>
+ P2<=
/span>
+ P3<=
/span> +
……
•#37
Determine the total pressure of a gas mixture that contains oxygen, nitrogen, and helium if the parti=
al pressures of the gases are as follows: Pox=
ygen =3D =
20.0kPa, Pnitrogen 46.7 kPa, and Phe=
lium =3D<=
span
style=3D'mso-spacerun:yes'> 26.7 kPa.
=
•#37
Determine the total pressure of a gas mixture that contains oxygen, nitrogen, and helium if the parti=
al pressures of the gases are as follows: Pox=
ygen =3D =
20.0kPa, Pnitrogen 46.7 kPa, and Phe=
lium =3D<=
span
style=3D'mso-spacerun:yes'> 26.7 kPa.
PT =3D P1
+ P2<=
/span>
+ P3<=
/span> +
……
PT =3D Poxygen + P<=
/span>ni=
trogen + P<=
/span>he=
lium =
3;
PT =3D 20.0 k Pa +
45.7 k Pa + 26.7 k Pa
PT =3D 92.4 k Pa
Graha=
m’s
Law of Effusion
•The rate of effusion of a gas is inversely proport=
ional =
to the square root of the gas’s molecular
mass.
Diffusion
– mixing of gas
Effusion
– gas escapes through a tiny hole in a container.
Rate A<=
/span>
=3D ͩ=
0;molar mass B
Rate B<=
/span>
√molar mass A&=
#13;
![](3D"slide0057_image036.gif")
![](3D"slide0057_image037.gif")
![](3D"slide0057_image038.gif")
•Compare the rate of effusion of the air component =
nitrogen (molar mass =3D28 g/mol) and helium (mola=
r mass =3D 4.0 g/mol)
•Compare the rate of effusion of the air component =
nitrogen (molar mass =3D28 g/mol) and helium (mola=
r mass =3D 4.0 g/mol)
=
;
• =
;
Rate A =3D ͩ=
0;molar mass B
• Rate B<=
/span>
√molar mass A&=
#13;
• Rate
He =3D ͩ=
0; 28.=
0 =3D=
5.3 =3D 2.7
• Rate N2=
√ 4.0
=3D =
2
•
• Helium ef=
fuses
2.7 times faster than nitrogen
•
![](3D"slide0073_image039.gif")
![](3D"slide0073_image040.gif")
![](3D"slide0073_image041.gif")
![](3D"slide0073_image042.gif")
![](3D"slide0073_image043.gif")
![](3D"slide0073_image044.gif")
![](3D"slide0073_image045.gif")
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